Calculate the value of the equilibrium constant for the following reaction: PbI2 (s) + 2I- (aq) <-> PbI4 ^ 2- (aq) The solubility product constant Ksp for PbI2 is 9.8e-9 and the formation constant Kf for PbI4^2- is 3.0e4

Question

I have Ksp=[PbI4^2-]
Kf= [PbI4^2-] / [PbI2][I-]^2
Kc= [PbI4^2-] / [I-]^2

I don't know where to go from here

Answer 1

You need to re-write the equations as such.
PbI2(s) ==> Pb^2+ + 2I^-
Pb^2+ + 4I^- --> [PbI4]^2-
--------------------------
add the two equation to get this.
PbI2(s) + 2I^- ==? [PbI4]^2- for which
Keq = Ksp*Kf

You can plug in Ksp and Kf and see that Keq is what it should be.
Ksp = (Pb^2+)(I^-)^2
Kf = [PbI4^2-]/(Pb^2+)(I^-)^4
Keq = [PbI4^2-]/(I^-)^2

Answer 2

thank you!!!!